Question

sin(arccos√3/3+arcsin(-1/2))
помогитеееее

Answer 1

sin(arccos√3/3 +arcsin(-1/2))  ;
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sin(α +β) =sinα*cosβ +cosα*sinβ   [α = arccos√3/3 ;β =arcsin(-1/2) ]
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sin(arccos√3/3 +arcsin(-1/2))= sin(arccos√3/3)*cos(arcsin(-1/2)) + cos(arccos√3/3)*sin(arcsin(-1/2))  =
sin(arcsin√6/3)*cos(arcccos(√3/2)+√3/3*(-1/2)
=√6/3*√3/2 - √3)/6 =(3√2 -√3)/6 .